In the rhombus ABCD, AB=10 and AC=12. Find the area of the rhombus.
I am confused about how to solve this problem.
let E be the intersection of the diagonals AC andBD
In a rhombus the diagonals right-bisect each other
so triangle ABE has
AB=12 , AE = 6 , and by Pythagoras BE = 8
Area of rhomus = 1/2 of the product of its diagonals
= (1/2)(12)(16) = 96
How do I know the value of the diagonals?
You told me one was 12 , AC = 12
and I found half of the other for you
I said BE = 8, so BD = 16
Ok, I understand that. But I am confused as to how you found the second diagonal (BD)?
To find the area of a rhombus, you can use the formula:
Area = (diagonal_1 * diagonal_2) / 2
In this case, the diagonals of the rhombus are not given directly. However, we can find them using the given information.
Let's start by drawing the rhombus ABCD:
A
/ \
/ \
/ \
/ \
B ------- C
\ /
\ /
\ /
\ /
D
We know that AB = 10 and AC = 12. Since a rhombus has equal sides, we can also say AD = BC = 10, and BD = AC = 12.
Now, we can observe that the diagonals of a rhombus intersect each other at right angles and bisect each other. Let's name the point where the diagonals intersect as E.
A
/ \
/ \
/ E \
/-------\
B ------- C
\ /
\ /
\ /
\ /
D
Since the diagonals bisect each other, we can say that AE = EC = 12/2 = 6 and BE = ED = 10/2 = 5.
Now, we have the lengths of the diagonals. We can substitute these values into the formula for the area of a rhombus:
Area = (diagonal_1 * diagonal_2) / 2
= (BE * AC) / 2
= (5 * 12) / 2
= 60 / 2
= 30 square units
So, the area of the rhombus ABCD is 30 square units.